Simplify and expand the following expression: $ \dfrac{1}{2t + 18}- \dfrac{4}{t - 5}+ \dfrac{t}{t^2 + 4t - 45} $
Explanation: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $2$ out of denominator in the first term: $ \dfrac{1}{2t + 18} = \dfrac{1}{2(t + 9)}$ We can factor the quadratic in the third term: $ \dfrac{t}{t^2 + 4t - 45} = \dfrac{t}{(t + 9)(t - 5)}$ Now we have: $ \dfrac{1}{2(t + 9)}- \dfrac{4}{t - 5}+ \dfrac{t}{(t + 9)(t - 5)} $ The least common multiple of the denominators is: $ 2(t + 9)(t - 5)$ In order to get the first term over $2(t + 9)(t - 5)$ , multiply by $\dfrac{t - 5}{t - 5}$ $ \dfrac{1}{2(t + 9)} \times \dfrac{t - 5}{t - 5} = \dfrac{t - 5}{2(t + 9)(t - 5)} $ In order to get the second term over $2(t + 9)(t - 5)$ , multiply by $\dfrac{2(t + 9)}{2(t + 9)}$ $ \dfrac{4}{t - 5} \times \dfrac{2(t + 9)}{2(t + 9)} = \dfrac{8(t + 9)}{2(t + 9)(t - 5)} $ In order to get the third term over $2(t + 9)(t - 5)$ , multiply by $\dfrac{2}{2}$ $ \dfrac{t}{(t + 9)(t - 5)} \times \dfrac{2}{2} = \dfrac{2t}{2(t + 9)(t - 5)} $ Now we have: $ \dfrac{t - 5}{2(t + 9)(t - 5)} - \dfrac{8(t + 9)}{2(t + 9)(t - 5)} + \dfrac{2t}{2(t + 9)(t - 5)} $ $ = \dfrac{ t - 5 - 8(t + 9) + 2t} {2(t + 9)(t - 5)} $ Expand: $ = \dfrac{t - 5 - 8t - 72 + 2t}{2t^2 + 8t - 90} $ $ = \dfrac{-5t - 77}{2t^2 + 8t - 90}$